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Computing Matrix, finding kernel and image

by specialnlovin
Tags: bases, image, kernel, linear algebra
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specialnlovin
#1
Nov2-10, 11:22 AM
P: 19
Let T: R[x]2[tex]\rightarrow[/tex] R[x]3 be defined by T(P(x))=xP(x). Compute the matrix of x with respect to bases {1,x,x2} and {1,x,x2,x3}. Find the kernel and image of T.

I know how to do this when given bases without exponents, however I do not know exactly what this is saying and therefore am having a hard time starting it.
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micromass
#2
Nov2-10, 11:59 AM
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Let's compute the first column.

You take the first basis element, that is 1. Now you'll have to express T(1)=x in terms of the basis {1,x,x,x}. That would be (0,1,0,0). So the first column would consist out of

[tex] \left(\begin{array}{ccc}
0 & ? & ? \\
1 & ? & ? \\
0 & ? & ? \\
0 & ? & ?
\end{array} \right) [/tex]

Now for the second and third column, you'll have to express T(x) and T(x) in terms of the basis {1,x,x,x}.
specialnlovin
#3
Nov2-10, 03:15 PM
P: 19
Okay so matrix T(x)=x2 and with respect to the basis {1,x,x,x} the second column would be (0,0,1,0), T(x2)=x3 and with respect to the basis {1,x,x,x} would be (0,0,0,1).
I would then say that
T=(0 0 0)
(1 0 0)
(0 1 0)
(0 0 1)
with respect to {1,x,x,x} right? Not with respect to {1,x,x} and {1,x,x,x}
so then to find the im(T) I just use
(0 0 0)
(1 0 0)
(0 1 0)
(0 0 1)
and solve it with respect to the basis {1,x,x,x} right?
Then how would I go about solving for the ker(T)?

micromass
#4
Nov2-10, 03:24 PM
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Computing Matrix, finding kernel and image

For the Ker(T). Take your matrix A. Then wonder when Ax=0. This you should be able to solve easily (it's a system with 3 unknowns and 4 equations)
micromass
#5
Nov2-10, 03:34 PM
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So, if (x,y,z) is in the kernel, then you must have

[tex]
\left( \begin{array}{ccc}
0 & 0 & 0\\
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array} \right) \left( \begin{array}{c}
x\\
y\\
z
\end{array} \right) = 0
[/tex]

It's not hard to see that this can only be the case iff x=y=z=0
specialnlovin
#6
Nov2-10, 03:45 PM
P: 19
right, that just seemed way too easy I thought I was doing it wrong.
Then the im(T) is the span of e2, e3, and e4
micromass
#7
Nov2-10, 03:47 PM
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P: 18,331
Yes, your image is correct to.

And remember, math doesnt have to be difficult


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